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integrate from 1 to 0 for 2/ (1+x^2) dx
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\[\int\limits_1^0\frac{2}{1+x^2}\text dx\] did you say the limits in the right order
yeah i did its from 0 to 1
\[\int\limits_0^1\frac{2}{1+x^2}\text dx\]
correct
\[=2\int\limits_{0}^{1} 1/(1+x^2) dx\]\[=[2\tan^{-1} x]_{0}^{1}\]\[=2\tan^{-1} 1 - 2\tan^{-1} 0\]
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\[=2(\pi/4) -0\]\[=\pi/2\]
why is it tan?
the integration of 1/(1+x^2) with respect to x is tan^-1 x not tan x.
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